MrEngenic (enthusiast) 09/20/05 12:52 PM
 Re: Standard Deviation??

Quote:

Put another way: if you have a winrate of 2BB/100 and a SD of 16BB/100, there is about a 68% chance that, in your next 100 hands, you will win between -14BB and 18BB.

Quote:

Put another way: there is a 95% chance that, in your next 100 hands, you will win between -30BB and 34BB.

Not to get picky, but these are common misconceptions in the poker world. Allow me to elaborate...

I am pretty sure that the winnings for each 100 hands does NOT have a normal distribution. I have not checked it because PT won't allow me to export vectors of total winnings for each 100th hand, so I don't know it for a fact though. I suspect that the distribution would have heavier tails than the normal distribution and it's probably not symmetric.

However, you can use the normal distribution to calculate a confidence interval for your true win rate. But this is because if you pick many samples from ANY distribution, the mean will be normally distributed according to the Central Limit Theorem.

let X1, X2, X3... Xn be your winnings for 100 hands each with an UNKNOWN distrubution, and say that the true expcted value is "my", the sample mean is Xbar and the standard deviation is "sigma".

Then (Xbar-my)/(sigma*sqrt(n)) will be normally distributed and a confidence interval for my will be Xbar+-z(a/2)*sigma*sqrt(n) with a confidence level of APPROXIMATELY 100*(1-a).

The confidence interval is defined as: Take a sample from the distribution specified by your estimated variables ,my and sigma, and calculate a confidence interval based on your sample and chosen a. 100*(1-a) % of the time you do this the confidence interval will cover your my.

But it was a good analysis coming from a bio major. Close enough.